Given the equilibrium constant: $\mathrm{K}_{\mathrm{C}}$ of the reaction: $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}+(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2}+(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ is $10 \times 10^{15=\text {, calculate the }} \mathrm{E}_{\text {cell }}^{\theta}$ of the reaction of $298 \mathrm{~K}\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}}\right.$ at $\left.298 \mathrm{~K}=0.059 \mathrm{~V}\right]$

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Redox Reactions and Electrochemistry

Given the equilibrium constant:

K_{C}

of the reaction:

Cu (s) + 2 Ag + (aq) \to Cu^{2} + (aq) + 2 Ag (s)

10 \times 1 0^{15 =, calculate the} E_{cell}^{θ}

of the reaction of

298 K [2.303 \frac{RT}{F}

298 K = 0.059 V]

Answer: (C)

At equilibrium

Δ G^{\circ} = - n F E_{cell}^{\circ} = - 2.303 R T lo g K

, so

E_{cell}^{\circ} = \frac{2.303 R T}{n F} lo g K

. With two electrons transferred and

K = 1 0^{16}

, and using

\frac{2.303 R T}{F} = 0.059

E_{cell}^{\circ} = \frac{0.059}{2} lo g 1 0^{16} = 0.0295 \times 16 \approx 0.47

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option